A LOWER BOUND FOR THE BERGMAN KERNEL
AND THE BOURGAIN-MILMAN INEQUALITY
ZBIGNIEW BLOCKI
Abstract. For pseudoconvex domains in Cn we prove a sharp lower
bound for the Bergman kernel in terms of volume of sublevel sets of the
pluricomplex Green function. For n = 1 it gives in particular another
proof of the Suita conjecture. If Ω is convex then by Lempert’s theory the estimate takes the form KΩ (z) ≥ 1/λ2n (IΩ (z)), where IΩ (z) is
the Kobayashi indicatrix at z. One can use this to simplify Nazarov’s
proof of the Bourgain-Milman inequality from convex analysis. Possible
further applications of Lempert’s theory in this area are also discussed.
1. Introduction
For a domain Ω in Cn and w ∈ Ω we are interested in the Bergman kernel
Z
2
KΩ (w) = sup{|f (w)| : f ∈ O(Ω),
|f |2 dλ2n ≤ 1}
Ω
and in the pluricomplex Green function with pole at w
GΩ,w = sup{u ∈ P SH − (Ω) : lim sup(u(z) − log |z − w|) < ∞}.
z→w
Our main result is the following bound:
Theorem 1. Assume that Ω is pseudoconvex. Then for w ∈ Ω and a ≥ 0
we have
1
(1)
KΩ (w) ≥ 2na
.
e λ2n ({GΩ,w < −a})
This estimate seems to be very accurate. It is certainly optimal in the
sense that if Ω is a ball centered at w then we get equality in (1) for all a.
It is useful and not trivial already for n = 1. Note that in this case if we let
a tend to ∞ then we immediately obtain
1
(2)
KΩ ≥ c2Ω ,
π
where
cΩ (z) = exp(lim (GΩ,z (ζ) − log |ζ − z|))
ζ→z
is the logarithmic capacity of C \ Ω with respect to z. This is precisely the
inequality conjectured by Suita [17] and recently proved in [3].
1
2
Z. BLOCKI
¯
Our proof of Theorem 1 uses the L2 -estimate for the ∂-equation
of Donnelly-Fefferman [7] from which we can first get a weaker version of (1):
KΩ (w) ≥
c(n, a)
,
λ2n ({GΩ,w < −a})
where
c(n, a) =
Ei (na)
Ei (na) + 2
2
and
Z
∞
ds
ses
b
(for b > 0). Then we employ the tensor power trick and use the fact that
(3)
Ei (b) =
lim c(nm, a)1/m = e−2na .
m→∞
This way we get an optimal constant in (1).
Our new proof of the one-dimensional estimate (2) makes crucial use of
many complex variables. The use of the tensor power trick here replaces a
special ODE in [3]. It should be noted though that this works only for the
Suita conjecture, we do not get the Ohsawa-Takegoshi extension theorem
from Theorem 1.
It is probably interesting to investigate the limit of the right-hand side of
(1) as a tends to ∞ also in higher dimensions. We suspect that it always
exists, at least for sufficiently regular domains. This way we would get
a certain counterpart of logarithmic capacity in higher dimensions. Using
Lempert’s theory [13], [14] one can check what happens with this limit for
smooth and strongly convex domains, see Proposition 3 below. This way we
get the following bound:
Theorem 2. Let Ω be a convex domain in Cn . Then for w ∈ Ω
1
,
KΩ (w) ≥
λ2n (IΩ (w))
where
IΩ (w) = {ϕ0 (0) : ϕ ∈ O(∆, Ω), ϕ(0) = w}
is the Kobayashi indicatrix (here ∆ denotes the unit disc).
One can use Theorem 2 to simplify Nazarov’s approach [15] to the Bourgain-Milman inequality [6]. For a convex symmetric body (i.e. open, bounded) L in Rn its dual is given by
L0 := {y ∈ Rn : x · y ≤ 1 for all x ∈ L}.
The product λn (L)λn (L0 ) is called a Mahler volume of L. It is independent
of linear transformations and on an inner product in Rn , and thus depends
A LOWER BOUND FOR THE BERGMAN KERNEL
3
only on the finite dimensional Banach space structure whose unit ball is L.
The Blaschke-Santaló inequality says that the Mahler volume is maximized
by balls.
On the other hand, the still open Mahler conjecture states that it is
minimized by cubes. A partial result in this direction is the BourgainMilman inequality [6] which says that there exists c > 0 such that
4n
(4)
λn (L)λn (L0 ) ≥ cn .
n!
The Mahler conjecture is equivalent to saying that we can take c = 1 in (4).
Currently, the best known constant in (4) is π/4 and is due to Kuperberg
[12].
Nazarov [15] recently proposed a complex-analytic approach to (4). He
considered tube domain TL := L + iRn and proved the following bounds for
the Bergman kernel at the origin:
(5)
(6)
KTL (0) ≤
KTL (0) ≥
n! λn (L0 )
π n λn (L)
π 2n
4
1
.
(λn (L))2
(π/4)3 .
This gave (4) with c =
The upper bound (5) was obtained relatively
easily from Rothaus’ formula for the Bergman kernel in tube domains (see
[16] and [10]):
Z
1
dλn
KTL (0) =
,
(2π)n Rn JL
where
Z
JL (y) =
e−2x·y dλn (x).
L
¯
For the lower bound (6) Nazarov used the Hörmander estimate [9] for ∂.
We will show that (6) follows easily from Theorem 2. It should be noted
however that although we are using the Donnelly-Fefferman estimate here,
it can be deduced quite easily from the Hörmander estimate (see [1]), so the
latter still plays a crucial role.
We conjecture that in fact the following lower bound holds:
π n
1
(7)
KTL (0) ≥
.
4
(λn (L))2
Since we have equality for cubes, this would be optimal. In Section 4 we
discuss possible applications of Lempert’s theory to this problem.
The author learned about the Nazarov paper [15] from professor Vitali
Milman during his visit to Tel Aviv in December 2011. He is also grateful
to Semyon Alesker for his invitation and hospitality.
4
Z. BLOCKI
2. Proofs of Theorems 1 and 2
Proof of Theorem 1. By approximation we may assume that Ω is bounded
and hyperconvex. We may also assume that a > 0, as for a = 0 it is enough
to take f ≡ 1. Denote G := GΩ,w . We will use the Donnelly-Fefferman
estimate [7] (see also [1] and [2]) with
ϕ := 2nG,
ψ := − log(−G)
and
¯ ◦ G) = χ0 ◦ G ∂G,
¯
α := ∂(χ
where χ will be determined later. We have
¯ ≤ G2 (χ0 ◦ G)2 i∂ ∂ψ.
¯
iᾱ ∧ α ≤ (χ0 ◦ G)2 i∂G ◦ ∂G
¯ = α and
We will find u ∈ L2loc (Ω) with ∂u
Z
Z
Z
2
2 −ϕ
|u| dλ2n ≤
|u| e dλ2n ≤ C
G2 (χ0 ◦ G)2 e−2nG dλ2n ,
Ω
Ω
Ω
where C is an absolute constant (in fact, the optimal one is C = 4, see
[2, 4]). We now set
(
0
t ≥ −a,
χ(t) := R −t e−ns
s ds, t < −a,
a
so that
Z
|u|2 dλ2n ≤ C λ2n ({G < −a}).
Ω
Since e−ϕ is not integrable near w, we have u(w) = 0. Therefore holomorphic
f := χ ◦ G − u satisfies
f (w) = χ(−∞) = Ei (na)
with Ei given by (3). We also have (with || · || denoting the L2 -norm in Ω)
√ p
||f || ≤ ||χ ◦ G|| + ||u|| ≤ (χ(−∞) + C) λ2n ({G < −a}).
Therefore
KΩ (w) ≥
|f (w)|2
c(n, a)
,
≥
||f ||2
λ2n ({G < −a})
where
Ei (na)2
√
.
(Ei (na) + C)2
We are now going to use the tensor power trick. For big natural m consider
e = Ωm in Cnm and w
e Then
the domain Ω
e = (w, . . . , w) ∈ Ω.
c(n, a) =
KΩe (w)
e = (KΩ (w))m
A LOWER BOUND FOR THE BERGMAN KERNEL
5
and by [11] (see also [8])
GΩ,
(z 1 , . . . , z m ) = max G(z j ),
ew
e
j=1,...,m
therefore
λ2nm ({GΩ,
< −a}) = (λ2n ({G < −a})m .
ew
e
It follows from the previous part that
(KΩ (w))m ≥
c(nm, a)
(λ2n ({G < −a}))m
and it is enough to use the fact that
lim c(nm, a)1/m = e−2na .
m→∞
Theorem 2 follows immediately from Theorem and the following result by
approximation.
Proposition 3. Assume that Ω is a bounded, smooth, strongly convex domain in Cn . Then for any w ∈ Ω
(8)
lim e2na λ2n ({GΩ,w < −a}) = λ2n (IΩ (w)).
a→∞
Proof. Denote I := IΩ (w), G := GΩ,w , we may assume that w = 0. By the
results of Lempert [13] there exists a diffeomorphism Φ : I¯ → Ω̄ such that
for v ∈ ∂I the mapping ∆ 37→ Φ(ζv) is a geodesic in Ω, that is
(9)
G(Φ(ζv)) = log |ζ|.
(Φ can be treated as an exponential map for the Kobayashi distance.) We
also have
Φ(ζv) = ζv + O(|ζ|2 ).
By (9)
{G < −a} = Φ(e−a int I)
and therefore
Z
λ2n ({G < −a}) =
e−a I
Since Φ0 (0) is the identity, we obtain (8).
Jac Φ dλ2n .
6
Z. BLOCKI
3. Applications to the Bourgain-Milman Inequality
Assume that L is a convex symmetric body in Rn . In view of Theorem
2, in order to prove Nazarov’s lower bound (6) it is enough to show the
estimate
2n
4
(10)
λ2n (ITL (0)) ≤
(λn (L))2 .
π
But this follows immediately from the following:
Proposition 4. ITL (0) ⊂
4
(L̄ + iL̄).
π
Proof. We will use an idea of Nazarov [15] here. Let Φ be a conformal
mapping from the strip {|Re ζ| < 1} to ∆ with Φ(0) = 0, so that |Φ0 (0)| =
π/4. For u ∈ L0 we can then define F ∈ O(Ω, ∆) by F (z) = Φ(z · u). For
ϕ ∈ O(∆, Ω) with ϕ(0) = 0 by the Schwarz lemma we have |(F ◦ ϕ)0 (0)| ≤ 1.
Therefore |ϕ0 (0) · u| ≤ 4/π and π4 ITL (0) ⊂ LC , where
LC = {z ∈ Cn : |z · u| ≤ 1 for all u ∈ L0 } ⊂ L̄ + iL̄.
It will be convenient to use the notation JL :=
proof of Proposition 4
π
4 ITL (0),
so that by the
JL ⊂ LC ⊂ L̄ + iL̄.
(11)
We thus have λ2n (JL ) ≤ (λn (L))2 but we conjecture that
π n
(12)
λ2n (JL ) ≤
(λn (L))2 .
4
Note that J[−1,1]n = ∆n , so that we have equality for cubes. The inequality
(12) would give the optimal lower bound for the Bergman kernel in tube
domains (7).
We first give an example that (11) cannot give us (12):
Example. Let L = {x21 + x22 < 1} be the unit disc in R2 . One can then show
that LC = {|z|2 ≤ 1 + (x1 y2 − x2 y1 )2 } and
λ4 (LC ) =
π 4 π 2
2π 2
>
=
(λ2 (L))2 .
3
16
4
A LOWER BOUND FOR THE BERGMAN KERNEL
7
4. Lempert’s Theory in Tube Domains
Our goal is to approach (12) using Lempert’s theory. First assume that Ω
is bounded, smooth, strongly convex domain in Cn . Then for any z, w ∈ Ω,
¯ Ω̄) such that
z 6= w, there exists unique extremal disc ϕ ∈ O(∆, Ω) ∩ C ∞ (∆,
ϕ(0) = w, ϕ(ξ) = z for some ξ with 0 < ξ < 1, and
GΩ,w (ϕ(ζ)) = log |ζ|,
ζ ∈ ∆.
Lempert [13] showed in particular the following characterization of extremal
¯ Ω̄) is extremal if and only if ϕ(∂∆) ⊂ ∂Ω
discs: a disc ϕ ∈ O(∆, Ω) ∩ C(∆,
¯ Cn ) such that the vector eit h(eit ) is
and there exists h ∈ O(∆, Cn ) ∩ C(∆,
outer normal to ∂Ω at ϕ(eit ) for every t ∈ R.
Lempert [13] also proved that for every extremal disc ϕ in Ω there exists
a left-inverse F ∈ O(Ω, ∆) (that is F (ϕ(ζ)) = ζ for ζ ∈ ∆). It solves the
equation
(13)
(z − ϕ(F (z))) · h(F (z)) = 0,
z ∈ Ω.
Now assume that L is a smooth, strongly convex body in Rn . Although
TL is neither bounded nor strongly convex, we may nevertheless try to apply
Lempert’s condition for extremal discs (the details have been worked out by
n
¯ n
Zajac
, [18]). First note that h ∈ O(∆, C ) ∩ C(∆, C ) in our case must
satisfy
(14)
Im (e−it h(eit )) = 0,
t ∈ R.
It follows that h must be of a very special form:
¯ satisfies (14) then h(ζ) = a + bζ + āζ 2
Lemma 5. ([18]) If h ∈ O(∆) ∩ C(∆)
for some a ∈ C and b ∈ R.
Proof. Set a := h(0). Then for ζ ∈ ∂∆
h(ζ)
h(ζ) − a
0 = Im
= Im
− āζ
ζ
ζ
and therefore
h(ζ) − a
− āζ = b ∈ R,
ζ
¯
ζ ∈ ∆.
We thus see that in our case h must be of the form
h(ζ) = w + ζb + ζ 2 w̄,
¯
ζ ∈ ∆,
8
Z. BLOCKI
for some w ∈ Cn and b ∈ Rn . Then the extremal disc ϕ for TL associated
with h satisfies
b + Re (e−it w)
it
−1
(15)
Re ϕ(e ) = ν
,
|b + Re (e−it w)|
where
ν : ∂L → Sn−1
is the Gauss map.
¯ we can recover the values of ϕ in ∆ from the values
For ϕ ∈ O(∆) ∩ C(∆)
of Re ϕ on ∂∆ using the Schwarz formula:
Z 2π it
1
e +ζ
ϕ(ζ) =
Re ϕ(eit ) dt + iIm ϕ(0), ζ ∈ ∆.
2π 0 eit − ζ
Therefore extremal discs satisfying (15) are given by
Z 2π it
1
e + ζ −1 b + Re (e−it w)
ϕ(ζ) =
dt + iIm ϕ(0),
ν
2π 0 eit − ζ
|b + Re (e−it w)|
ζ ∈ ∆.
We now assume that L is in addition symmetric and then consider the case
when b = 0 and Im ϕ(0) = 0:
Z 2π it
1
e + ζ −1 Re (e−it w)
(16)
ϕ(ζ) =
ν
dt.
2π 0 eit − ζ
|Re (e−it w)|
Since L is symmetric the function B(t) under the integral in (16) satisfies
B(t + π) = −B(t). We thus have ϕ(0) = 0 and one can show (see [18]
for details) that all geodesics of TL passing through the origin are given by
(16). They are bounded and smooth up to the boundary if Re w and Im w
are linearly independent in Rn . If Re w and Im w are linearly dependent
(and w 6= 0) then (16) gives special extremal discs of the form
ϕ(ζ) = Φ−1 (ζ) x,
x ∈ ∂L,
where Φ is as in the proof Proposition 4. Left-inverses to these ϕ are then
given by F (z) = Φ(z · u) for unique u ∈ ∂L0 with x · u = 1.
For geodesics (16) we have
Z
1 2π it −1 Re (eit w)
e ν
dt.
(17)
ϕ0 (0) =
π 0
|Re (eit w)|
These vectors parametrize the boundary of the Kobayashi indicatrix IL (0).
If F ∈ O(Ω, ∆) is the left-inverse of ϕ satisfying (13) we get, since h0 (0) = 0,
w
F 0 (0) = 0
.
ϕ (0) · w
Therefore
(18)
JL = {z ∈ Cn : |z · w| ≤ |Ψ(w)| for all w ∈ (Cn )∗ },
A LOWER BOUND FOR THE BERGMAN KERNEL
9
where
Z
1 2π it
Re (eit w)
−1
Ψ(w) =
e w·ν
dt.
4 0
|Re (eit w)|
Both (17) and (18) give a description of the set JL . It would be interesting
to try to use it to prove (12). We can at least show this for a ball:
Example. Let B = {|x| < 1} be the unit ball in Rn . For w ∈ (Cn )∗ we have
Z
Z
1 2π Im (eit w) · Re (eit w)
1 2π d
Im Ψ(w) =
dt
=
−
|Re (eit w)|dt = 0
4 0
|Re (eit w)|
4 0 dt
and thus
Z
1 2π
π
Ψ(w) =
|Re (eit w)|dt ≤ √ |w|.
4 0
8
√
By (18) JB is contained in a ball with radius π/ 8 in Cn . Therefore
λ2n (JB ) ≤
π 3n
.
8n n!
On the other hand,
π n/2
,
Γ( n2 + 1)
and we see that (12) holds for B if n ≥ 3. To show this also for n = 2
we have to use in addition Proposition 4: JB ⊂ (B̄ + iB̄) ∩ (r0 B̄), where
p
√
r0 = π/ 8. With ρ0 = r02 − 1 we will get
Z 1
π 4 π 2
π6 π4 π2
2 2
2
+ −
<
=
λ4 (JB ) ≤ π ρ0 +π
ρ(r02 −ρ2 )dρ =
(λ2 (B))2 .
256
16
2
16
4
ρ0
λn (B) =
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10
Z. BLOCKI
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,
Uniwersytet Jagielloński, Instytut Matematyki, Lojasiewicza 6, 30-348 Kraków, Poland
E-mail address: [email protected], [email protected]